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  1. Indestructibility and the linearity of the Mitchell ordering.Arthur W. Apter - 2024 - Archive for Mathematical Logic 63 (3):473-482.
    Suppose that \(\kappa \) is indestructibly supercompact and there is a measurable cardinal \(\lambda > \kappa \). It then follows that \(A_0 = \{\delta is a measurable cardinal and the Mitchell ordering of normal measures over \(\delta \) is nonlinear \(\}\) is unbounded in \(\kappa \). If the Mitchell ordering of normal measures over \(\lambda \) is also linear, then by reflection (and without any use of indestructibility), \(A_1= \{\delta is a measurable cardinal and the Mitchell ordering of normal measures (...)
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  • Indestructibility and destructible measurable cardinals.Arthur W. Apter - 2016 - Archive for Mathematical Logic 55 (1-2):3-18.
    Say that κ\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$${\kappa}$$\end{document}’s measurability is destructible if there exists a κ\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$${\kappa}$$\end{document}. It then follows that A1={δ<κ∣δ\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$${A_{1} = \{\delta < \kappa \mid \delta}$$\end{document} is measurable, δ is not a limit of measurable cardinals, δ is not δ+ strongly compact, and δ’s measurability is destructible when forcing with partial orderings having rank below λδ} is unbounded (...)
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  • Indestructibility, measurability, and degrees of supercompactness.Arthur W. Apter - 2012 - Mathematical Logic Quarterly 58 (1):75-82.
    Suppose that κ is indestructibly supercompact and there is a measurable cardinal λ > κ. It then follows that A1 = {δ < κ∣δ is measurable, δ is not a limit of measurable cardinals, and δ is not δ+ supercompact} is unbounded in κ. If in addition λ is 2λ supercompact, then A2 = {δ < κ∣δ is measurable, δ is not a limit of measurable cardinals, and δ is δ+ supercompact} is unbounded in κ as well. The large cardinal (...)
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